Chapter 1: Real Numbers (Exercise 1.3)

Chapter 1: Real Numbers (Exercise 1.3)

Question 1:

Classify the following numbers as rational or irrational:

1. $2 - \sqrt{5}$

2. $(3 + \sqrt{23}) - \sqrt{23}$

3. $\frac{2\sqrt{7}}{7\sqrt{7}}$

4. $\frac{1}{\sqrt{2}}$

5. $2\pi$

Solution:

1. $2 - \sqrt{5}$:

   • Here, $2$ is a rational number and $\sqrt{5}$ is an irrational number.

   • The difference between a rational number and an irrational number is always irrational.

   • Answer: Irrational Number

2. $(3 + \sqrt{23}) - \sqrt{23}$:

   • Simplifying the expression: $3 + \sqrt{23} - \sqrt{23} = 3$

   • $3$ can be written in $\frac{p}{q}$ form as $\frac{3}{1}$.

   • Answer: Rational Number

3. $\frac{2\sqrt{7}}{7\sqrt{7}}$:

   • Cancelling $\sqrt{7}$ from numerator and denominator: $\frac{2}{7}$

   • $\frac{2}{7}$ is in $\frac{p}{q}$ form.

   • Answer: Rational Number

4. $\frac{1}{\sqrt{2}}$:

   • The quotient of a non-zero rational number and an irrational number is irrational.

   • Answer: Irrational Number

5. $2\pi$:

   • Here, $2$ is rational and $\pi$ is an irrational number.

   • The product of a non-zero rational number and an irrational number is irrational.

   • Answer: Irrational Number

Question 2:

Simplify each of the following expressions:

1. $(5 + \sqrt{7})(2 + \sqrt{5})$

2. $(5 + \sqrt{5})(5 - \sqrt{5})$

3. $(\sqrt{3} + \sqrt{7})^2$

4. $(\sqrt{11} - \sqrt{7})(\sqrt{11} + \sqrt{7})$

Solution:

1. $(5 + \sqrt{7})(2 + \sqrt{5})$:

   Using distributive law $(a + b)(c + d) = ac + ad + bc + bd$:

   $$= 5(2) + 5(\sqrt{5}) + \sqrt{7}(2) + \sqrt{7}(\sqrt{5})$$
   $$= 10 + 5\sqrt{5} + 2\sqrt{7} + \sqrt{35}$$

2. $(5 + \sqrt{5})(5 - \sqrt{5})$:

   Using the algebraic identity $(a + b)(a - b) = a^2 - b^2$:

   $$= (5)^2 - (\sqrt{5})^2$$
   $$= 25 - 5 = 20$$

3. $(\sqrt{3} + \sqrt{7})^2$:

   Using the algebraic identity $(a + b)^2 = a^2 + 2ab + b^2$:

   $$= (\sqrt{3})^2 + 2(\sqrt{3})(\sqrt{7}) + (\sqrt{7})^2$$
   $$= 3 + 2\sqrt{21} + 7$$
   $$= 10 + 2\sqrt{21}$$

4. $(\sqrt{11} - \sqrt{7})(\sqrt{11} + \sqrt{7})$:

   Using the algebraic identity $(a - b)(a + b) = a^2 - b^2$:

   $$= (\sqrt{11})^2 - (\sqrt{7})^2$$
   $$= 11 - 7 = 4$$

Question 3:

Rationalise the denominator of each of the following:

1. $\frac{1}{\sqrt{7}}$

2. $\frac{1}{\sqrt{7} - \sqrt{6}}$

3. $\frac{1}{\sqrt{5} + \sqrt{2}}$

4. $\frac{1}{\sqrt{7} - 2}$

Solution:

1. $\frac{1}{\sqrt{7}}$:

   Multiply numerator and denominator by rationalising factor $\sqrt{7}$:

   $$\frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{7}}{7}$$

2. $\frac{1}{\sqrt{7} - \sqrt{6}}$:

   Multiply numerator and denominator by rationalising factor $(\sqrt{7} + \sqrt{6})$:

   $$\frac{1}{\sqrt{7} - \sqrt{6}} \times \frac{\sqrt{7} + \sqrt{6}}{\sqrt{7} + \sqrt{6}} = \frac{\sqrt{7} + \sqrt{6}}{(\sqrt{7})^2 - (\sqrt{6})^2}$$
   $$= \frac{\sqrt{7} + \sqrt{6}}{7 - 6} = \frac{\sqrt{7} + \sqrt{6}}{1} = \sqrt{7} + \sqrt{6}$$

3. $\frac{1}{\sqrt{5} + \sqrt{2}}$:

   Multiply numerator and denominator by rationalising factor $(\sqrt{5} - \sqrt{2})$:

   $$\frac{1}{\sqrt{5} + \sqrt{2}} \times \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}} = \frac{\sqrt{5} - \sqrt{2}}{(\sqrt{5})^2 - (\sqrt{2})^2}$$
   $$= \frac{\sqrt{5} - \sqrt{2}}{5 - 2} = \frac{\sqrt{5} - \sqrt{2}}{3}$$

4. $\frac{1}{\sqrt{7} - 2}$:

   Multiply numerator and denominator by rationalising factor $(\sqrt{7} + 2)$:

   $$\frac{1}{\sqrt{7} - 2} \times \frac{\sqrt{7} + 2}{\sqrt{7} + 2} = \frac{\sqrt{7} + 2}{(\sqrt{7})^2 - (2)^2}$$
   $$= \frac{\sqrt{7} + 2}{7 - 4} = \frac{\sqrt{7} + 2}{3}$$

Question 4:

Find the values of $a$ and $b$ in the following equation:

Solution:

Rationalising the denominator of the Left Hand Side (LHS):

Expanding numerator using $(a+b)^2 = a^2 + 2ab + b^2$:

Equating LHS to RHS ($a + b\sqrt{6}$):

Comparing rational and irrational parts on both sides: